An electron microscope uses electrons, accelerated by a voltage of 50kV. Determine the de Broglie wavelength associated with the electrons. Taking other facters, such as numerical aperature etc. to same, how does the resolving power of an electron microscope compare with that of an optical microscope which uses yellow light.

Explanation

Kinetic energy of the electron,
`E = 50 keV = 50 xx 1.6 xx 10^(-16)J`
`:.` de Broglie wavelength,
`lamda = (h)/(sqrt(2mE))= (6.626 xx 10^(-34))/(sqrt(2 xx (9.1 xx 10^(-31)) xx (50 xx 1.6 xx 10^(-16)))) xx 10^(10) Å`
`= 0.055 Å`
So, `lamda lt ltlamda'`
The least distance (d) between two points that a microscope can observe separately is proportional to the wavelength of light used, i.e, `d prop lamda`.
As `lamda lt lt lamda'`, we have `d lt lt d'`. So an electron microscope can resolve two points very much closer. Then its resolving power is much higher than that of an optical microscope.